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    The point P with x-coordinate p lies on the curve y=x+c\x. Given that the gradient at P is -4 show that c=4p^2.
    Any help is appreciated
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    (Original post by toxi)
    The point P with x-coordinate p lies on the curve y=x+c\x. Given that the gradient at P is -4 show that c=4p^2.
    Any help is appreciated
    I did this question before but I can't seem to do it now, is the equation of the curve right? Because the solitary x is mucking me up.
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    y=x + (c/x)

    dy/dx= 1 - (c/x^2)

    you were told gradient = -4 which = dy/dx
    oops
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    my fault, it is y=1+c/x
    pls explain me how to do it....I dont wanna copy it i wanna understand it too
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    y = 1 + c/x

    dy/dx = - c/x²

    => -4 = - c/p² (by putting x coordinate p in)
    => -4p² = - c
    => c = 4p²
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    (Original post by toxi)
    my fault, it is y=1+c/x
    pls explain me how to do it....I dont wanna copy it i wanna understand it too
    OK:

    y = 1+c/x, then dy/dx = -c/x²

    At P, dy/dx = -4 and x = p, so substitute.

    -4 = -c/p²
    4 = c/p²
    4p² = c
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    I presume that y = (x+c)/x rather than y = x + (c/x).

    Then y = 1 + c/x so dy/dx = -c/x^2.

    Since dy/dx = -4 when x = p we have - c/p^2 = -4. So c = 4p^2.
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    lol yes he wrote the question incorrectly
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    got it....pretty easy in the end...thanx
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    (Original post by toxi)
    The point P with x-coordinate p lies on the curve y=x+c\x. Given that the gradient at P is -4 show that c=4p^2.
    Any help is appreciated
    y = (x + c)/x = 1 + cx^-1
    dy/dx = -cx^-2 = -c/x^2
    -4 = -c/x^2 = -c/p^2
    -4p^2 = -c
    c = 4p^2

    Too late
 
 
 

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